Problem: What is the slope of the line tangent to $f(x) = 2x^{2}-4x-5$ at $x = 2$ ?
The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}-4(x+h)-5) - (2x^{2}-4x-5)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})-4(x+h)-5) - (2x^{2}-4x-5)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}-4x-4h-5-2x^{2}+4x+5}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}-4h}{h}$ $ = \lim_{h \to 0} 4x+2h-4$ $ = 4x-4$ $ = (4)(2)-4$ $ = 4$